Question

A car of mass 2800 kg collides with a truck of mass 4000 kg, and just after the collision the car and truck slide along, stuck together, with no rotation. The car's velocity just before the collision was <36, 0, 0> m/s, and the truck's velocity just before the collision was <-15, 0, 29> m/s. (a) Your first task is to determine the velocity of the stuck-together car and truck just after the collision. (d) What is the increase in internal energy of the car and truck (thermal energy and deformation)?

Answer 1

Answer:

Final Velocity of the system = 18.07 m/s

Increase in the internal energy = 2836215.34 Joules

Explanation:

Treating the car and truck as a system we shall use the conservation of momentum to find the final velocities

Along the x axis the moumentum of each body is as under

$p_{cx}=m_{car}V_{x}\\\\p_{cx}=2800kg\times 36m/s\\\\p_{cx}=100800kgm/s\\\\p_{tx}=m_{truck}V'_{x}\\\\p_{tx}=4000kg\times -15m/s\\\\p_{tx}= -60000kgm/s$

Thus initial moumentum of the system along x direction is

$p_{cx}+p_{tx}=(100800-60000)kgm/s\\\\p_{initial}=40800kgm/s$

Now after collision since both the car and truck move as a single unit we have

$p_{final}=(m_{car}+m_{truck})v_{f}$

Equating the initial momentum with final momentum we get

$40800=(2800+4000)v_{fx}\\\\v_{fx}=\frac{40800}{6800}=6m/s$

Similarly we shall conserve the moumentum along the z direction also

$p_{car}=m_{car}\times v_{z}\\\\p_{car}=2800kg\times0m/s\\p_{car}=0\\\\p_{truck}=m_{truck}\times v_{z}\\p_{truck}=4000kg\times 29m/s\\\\p_{truck}=116000kgm/s\\\\p_{system}=116000kgm/s$

Equating initial and final momentum we have

$116000=(m_{car}+m_{truck})v'_{z}\\\\v'_{z}=\frac{116000}{2800+4000}\\\\v'_{z}=17.05m/s$

Thus the resultant velocity of the system after the collision is given by

$v_{res}=\sqrt{v_{x}^{2}+v_{z}^{2}}\\\\v_{res}=\sqrt{{6}^{2}+17.05^{2}}\\\\v_{res}=18.07m/s$

To find the increase in the internal energy we shall use the conservation of energy principle

Initial Energy =

$K.E_{car}+K.E_{truck}\\\\K.E_{car}=\frac{1}{2}m_{car}v_{car}^{2}\\\\K.E_{car}=\frac{1}{2}\times 2800kg\times (36m/s)^{2}\\\\K.E_{car}=1814400Joules\\\\K.E_{truck}=\frac{1}{2}m_{truck}v_{truck}^{2}\\\\K.E_{truck}=\frac{1}{2}\times 4000kg\times((-15m/s)^{2}+(29m/s)^{2})\\\\K.E_{truck}=2132000Joules\\\\K.E_{system}=(1814400+2132000)Joules=3946400Joules$

Similarly final energy of the system is

$K.E_{final}=\frac{1}{2}(m_{car}+m_{truck})v_{f}^{2}\\\\K.E_{final}=\frac{1}{2}(2800+4000)kg\times (18.07m/s)^{2}\\\\K.E_{final}=1110184.66Joules$

Thus the increase in the internal energy is the diffrence between initial and final energies

$\Delta IE_{system}=(3946400-1110184.66)joules\\\\\Delta IE_{system}=2836215.34Joules$

Answer 2

(a) The velocity of the car-truck system after collision is 18.08 m/s.

(b) The increase in internal energy of the car and truck is 2,835,031.2 J.

Velocity of the truck - car system after collision

The velocity of the system after collision is determined by applying the principle of conservation of linear momentum as shown below;

Final velocity in x - direction

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2800(36) + 4000(-15) = vx(2800 + 4000)

40,800 = 6800vx

vx = 6 m/s

Final velocity in z - direction

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2800(0) + 4000(29) = vz(2800 + 4000)

116,000 = 6800vz

vz = 17.06 m/s

Resultant velocity of the car-truck system after the collision $v= \sqrt{v_x^2 + v_z^2} \\\\v = \sqrt{6^2 + 17.06^2} \\\\v = 18.08 \ m/s$

Initial kinetic energy of the car and truck

K.E(car) = ¹/₂mv²

K.E(car) = ¹/₂ x (2800) x (36)²

K.E(car) = 1,814,400 J

v(truck) = √(15² + 29²) = 32.65

K.E(truck) = ¹/₂ x (4000) x (32.65)²

K.E(truck) = 2,132,045 J

K.E(total) =  1,814,400 J + 2,132,045 J = 3,946,445 J

Final kinetic energy of the system

K.E =  ¹/₂(m₁ + m₂)v²

K.E = ¹/₂ x (2800 + 4000) x (18.08)²

K.E = 1,111,413.8 J

Increase in internal energy

U = ΔK.E

U = 3,946,445 J - 1,111,413.8 J

U = 2,835,031.2 J

Learn more about conservation of linear momentum here: https://brainly.com/question/7538238